Maybe Gary or someone else has the answer to this one. I have 54*484^n+1 reserved. Clearly 484 = 22^2. A prime was found for 54*22^n+1, but I don't know what n it was prime for. If n was even then that would prove Sierpinski base 484.

[quote=rogue;208048]Maybe Gary or someone else has the answer to this one. I have 54*484^n+1 reserved. Clearly 484 = 22^2. A prime was found for 54*22^n+1, but I don't know what n it was prime for. If n was even then that would prove Sierpinski base 484.[/quote]
These are the primes for that base 22 k with n<=1000:
54*22^13+1 is 3-PRP! (0.0000s+0.0006s)
54*22^39+1 is 3-PRP! (0.0001s+0.0011s)
Both odd n.
It was eliminated so early that it's of no real value in the search for a 54*484^n+1 prime. If it had a prime with an even n, base 484 would have the same prime at half the n.

[quote=Batalov;208046]One last proof before going to sleep:

R414 is proven (k=46 = b/3[sup]2[/sup]); the rest is ditto.[/quote]
I was too sleepy last night, so I didn't elaborate. I will use this space to generalize.

Let b=k*x^2 and n=2m-1 is odd. Then
k*b[sup]2m-1[/sup]-1 = k*(k*x^2)[sup]2m-1[/sup]-1 = k[sup]2m[/sup]x[sup]2n[/sup] - 1[sup]2[/sup] = (k[sup]m[/sup]x[sup]n[/sup] - 1)(k[sup]m[/sup]x[sup]n[/sup] + 1), and is composite.
For the even n's, if there's a trivial factor (which is to be found case by case, using a hint from the srsieve and then doing modular arithmetics in mod 5, or mod 17, or mod N to be found), then the k is eliminated.

Here, b=414, x=3 (and k=46). And for even n, k*b[sup]2m[/sup]-1 [FONT=Times New Roman]≡[/FONT] 1*(-1)[sup]2m[/sup]-1 [FONT=Times New Roman]≡[/FONT] 0 (mod 5)

Other cases were ([I]may be typos here[/I]):
b=444, x=2
b=288, x=4 and "7/6" {x=7,y=6} (a variation to the above proof: 288=2^5*3^2, 392=2^3*7^2)
b=294, x=7 and "7/2" {x=7,y=2}
b=864, x=3 and x=12

Similar for k=b*x^2 (a special case of a multiple of base): left as an excercise.

In all cases, one thing is common: [B]k*b is a square[/B].
Ah, where were my eyes. :-) The whole thing is so easily re-written now:
Let k*b be a square, then for odd n's we trivially observe the difference of squares.
But I'll leave the blueprints. Could be educational. Sometimes such a simple idea comes only after a scribbled list... well, you know. Fun, fun.

Now, if [B]k*b^2 or k*b is a cube[/B], one obtains algebraics for both Riesel and Sierp for certain n's; similar (but rarer) for fifth degrees, etc.

Look for such cases in your bases.
__________

Now I'd like to get back to the earlier argument: should the sieve [I]or[/I] pfgw remove such cases by a fast factorization of [I]k[/I] and [I]b[/I]?
I think, both!
Or [I]the script[/I].
This is because when people start a new base, they initially use pfgw and [I]the script[/I]. They don't even get to the srsieve until much later.

[QUOTE=Mini-Geek;208050]These are the primes for that base 22 k with n<=1000:
54*22^13+1 is 3-PRP! (0.0000s+0.0006s)
54*22^39+1 is 3-PRP! (0.0001s+0.0011s)
Both odd n.
It was eliminated so early that it's of no real value in the search for a 54*484^n+1 prime. If it had a prime with an even n, base 484 would have the same prime at half the n.[/QUOTE]

I've searched up to n=~65000 for this with no luck yet. Considering how heavy this k/b combo is (>4.5% tests remain after sieving), I was hoping for a quick knock out.

Sierp 275, 281 & 307
 

Reserving Sierp 275, 281, 307 and 338 as new to n=25K

R405 is proven
 

R405 is proven with conj. k=146.
Data is attached.

Reserving as new S405 and R/S441.

54*484^69515+1 is prime.

At 186,639 digits, this will make it into the Prime Pages.

And t also proves the Sierpinski conjecture for base 484.

And it also removes a rather nasty conjecture with a single k remaining.

54*484^69515+1 is prime!

Riesel Base 285
 

2*285^1-1
4*285^71-1
6*285^1-1
8*285^2-1
10*285^2-1

With a conjectured k of 12, this conjecture is proven.

[quote=rogue;208443]54*484^69515+1 is prime![/quote]

Nice proof Mark. I wondered what that Sierp base 22 prime was that came across on top 5000. :smile: